# Expectation value of two annihilation operators

1. May 16, 2009

Hello,

I was studying about the effect of a beam splitter in a text on quantum optics. I understand that if a and b represent the mode operators for the two beams incident on the splitter, then the operator for one of the outgoing beams is the following,
$$c = \frac{(a + ib)}{\sqrt{2}}$$​

Now if I try to measure the intensity of this beam by a photodiode, the intensity will be proportional to-

$$<c^{\dag} c>$$

On evaluating this, I get,
$$\frac{\left(<a^{\dag} a> + <b^{\dag} b> + i(<a^{\dag} b> - <b^{\dag} a>)\right)}{2}$$
Now the book says, that this can be written as,
$$\frac{\left(<a^{\dag} a> + <b^{\dag} b> + i(<a^{\dag}><b> - <b^{\dag}><a>)\right)}{2}$$

I am unable to understand this step, that is $$<a^{\dag} b> = <a^{\dag}><b>$$
can someone please explain this.
I understand that these mode operators commute, but is this always true for any two commuting operators.

Thanks

2. May 16, 2009

### malawi_glenn

if you write what $< a^{\dagger} \, b >$ is, then I am sure that you can figure it out.